I've heard about people talking about the 1/N expansion as a mothod of nonperturbative QCD. This seems strange. In the 1/N expansion all you have is Feynman diagrams, which are by definition perturbative. Just resumming them to infinite orders doesn't really count as nonperturbative physics. Is it really jusified to think resummation has anything to do with noperturbative physics? Is there some subtle connection betweeen the two that I missed?

On one hand, the expansion (1) of $f$ yields an expansion that is zero to all orders perturbatively in $g_{YM}$. On the other hand, the expansion (2) of $f$ is exact already after the zeroth-order term in $1/N_c$.

@Vladimir Kalitvianski: 1/N expansion, just like ordinary $g_{YM}$ expansion, is an asymptotic series rather than a converging series. It's not finite in any sense.
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felixNov 18 '11 at 21:49

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@Qmechanic: My objection is that there are effects which are zero to all orders in perturbation theory, but only non-zero nonperturbatively. Baryon number violation for example. I'd like to correct your last sentence as "There are low-order perturbative terms in the latter expansion that correspond to all-order perturbative terms in the latter expansion, and vice versa, but there are no nonperturbative terms whatsoever." Am I wrong?
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felixNov 18 '11 at 21:53

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I updated the answer with a toy-example that is non-perturbative in the sense (1) but tree-level exact in the sense (2).
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Qmechanic♦Dec 19 '11 at 12:36